Let y=y(x) be the solution of the differentia | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given differential equation is $x\frac{dy}{dx}-y=x^2 \cot x$. Dividing by $x^2$, we get $\frac{x \frac{dy}{dx}-y}{x^2} = \cot x$. This can be

Vedclass · Accepted Answer

$-\pi$

Vedclass · Answer

(C) The given differential equation is $x\frac{dy}{dx}-y=x^2 \cot x$. Dividing by $x^2$, we get $\frac{x \frac{dy}{dx}-y}{x^2} = \cot x$. This can be written as $\frac{d}{dx}\left(\frac{y}{x}\right) = \cot x$. Integrating both sides with respect to $x$, we get $\frac{y}{x} = \int \cot x \, dx = \ln|\sin x| + C$. Given $y(\frac{\pi}{2}) = \frac{\pi}{2}$, we substitute $x = \frac{\pi}{2}$ and $y = \frac{\pi}{2}$: $\frac{\pi/2}{\pi/2} = \ln(\sin \frac{\pi}{2}) + C \implies 1 = \ln(1) + C \implies C = 1$. Thus, the solution is $y = x(\ln(\sin x) + 1)$. Now, calculate $y(\frac{\pi}{6}) = \frac{\pi}{6}(\ln(\sin \frac{\pi}{6}) + 1) = \frac{\pi}{6}(\ln(\frac{1}{2}) + 1) = \frac{\pi}{6}(1 - \ln 2)$. Calculate $y(\frac{\pi}{4}) = \frac{\pi}{4}(\ln(\sin \frac{\pi}{4}) + 1) = \frac{\pi}{4}(\ln(\frac{1}{\sqrt{2}}) + 1) = \frac{\pi}{4}(1 - \frac{1}{2}\ln 2)$. Finally, $6y(\frac{\pi}{6}) - 8y(\frac{\pi}{4}) = 6[\frac{\pi}{6}(1 - \ln 2)] - 8[\frac{\pi}{4}(1 - \frac{1}{2}\ln 2)]$. $= \pi(1 - \ln 2) - 2\pi(1 - \frac{1}{2}\ln 2) = \pi - \pi \ln 2 - 2\pi + \pi \ln 2 = -\pi$.

Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-y=x^{2}\cot x, x\in(0,\pi)$. If $y(\frac{\pi}{2})=\frac{\pi}{2}$, then $6y(\frac{\pi}{6})-8y(\frac{\pi}{4})$ is equal to :

Similar Questions

The general solution of the differential equation $y+\cos x(\frac{dy}{dx})-\cos^2 x=0$ is

The solution of the differential equation $\sqrt{1-y^2} dx + x dy - \sin^{-1} y dy = 0$ is

Let $f: [1, \infty) \to \mathbb{R}$ be a differentiable function defined as $f(x) = \int_1^x f(t) \, dt + (1 - x)(\log_e x - 1) + e$. Then the value of $f(f(1))$ is:

The solution of the differential equation $(x+2y^3) \frac{dy}{dx} = y$ is

An integrating factor of the differential equation $x \frac{dy}{dx} + y \log x = x e^x x^{-\frac{1}{2} \log x}$,$(x > 0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module